 Hello and welcome to this session. In this session we will discuss a question which says that two parts are drawn from a standard deck of parts without replacement. Find the probability of first part drawing a pin and then an ace. Second part drawing two aces. Now the first part in the solution of this question we should know are the result. And that is if A and B are dependent events then probability of event A and B is equal to probability of event A in the conditional probability of occurrence of event B given that event A occurs or we can write it as probability of event A intersection B is equal to probability of event A in the conditional probability of occurrence of event B given that event A occurs. Now this result we welcome to the key idea for solving the given question. Now let us start with the solution of the given question. Now we have given two parts are drawn from a standard deck of parts without replacement. Now in the first part we have to find the probability of drawing a pin and then an ace. Now we know that in a standard deck there are 52 cards where 26 cards are red in color and 26 cards are black in color and these 26 red cards consist of 13 hearts and 13 diamonds and these 26 black cards consist of 13 spades and 13 clubs and in a standard deck we have four tins and four aces. Now in the first part we have to find probability of drawing a pin and then an ace. So first we define the events. Now let event A as a pin is drawn and event B is an ace is drawn. So here we have to find probability of drawing a pin and then an ace which means we have to find probability of event A and B since two cards are drawn without replacement so the two draws are dependent events. Now using the result which is given to us in the key idea and B are dependent events then probability of event A and B is equal to probability of event A into conditional probability of occurrence of event B given that event A occurs that is probability of event A and B is equal to probability that a pin is drawn into probability of drawing an ace given that a pin is already drawn. Now we know that there are four tins in a deck of 52 cards so probability that a pin is drawn that is probability of event A is equal to number of outcomes favorable for event A that is forward upon total number of outcomes that is 52. Now pin is not replaced so remaining cards is equal to 52 cards minus one pin card that is equal to 51 cards. Now we want to draw an ace from remaining 51 cards. Now we have four aces so probability of drawing an ace given that a pin is already drawn is equal to four upon 51. So probability of event A and B is equal to probability of event A that is forward upon 52 into conditional probability of occurrence of event B given that event A occurs that is forward upon 51. Now we know that four into 13 is 52 so this is equal to four upon 13 into 51 that is 663 so this is the required answer for the first part. Now in the second part we have to find probability of drawing two aces. Now in the second part let us define the events. Now we have event A, ace is drawn and event B is again ace is drawn. Now we want to find probability of event A and B which is equal to probability of drawing an ace into probability of drawing an ace given that an ace is already drawn. Now we know that there are four aces in a deck of 52 cards so probability of drawing an ace is equal to four upon 52. Now four into 13 is 52 so this is equal to one upon 13 now when an ace is drawn then remaining cards is equal to 52 cards minus one ace card that is equal to 51 cards and since an ace is drawn so three aces are left. So probability of drawing an ace given that an ace is already drawn is equal to three upon 51 therefore probability of drawing two aces that is probability of event A and B is equal to one upon 13 into three upon 51 which is equal to now three into 17 is 51 so this is equal to one upon 13 into 17 that is 221. So this is the required answer for the second part and this is the solution of the given question. That's all for this session. Hope you all have enjoyed the session.